"Best" way to lower volume on a too-loud circuit?

[Bishop Vogue]

Madkatb asked this question a while back, but alas no replies. so I thought I'd pick up the torch since the stuff I'm designing these days regularly calls for this kind of a fix. Typically when I have a circuit that is great-sounding but much too loud, I'll just run a 10K-100K resistor to ground right at the output, before my volume pot.  But if there's one thing I know, it's that I know very little about circuit design.  So I'm wondering if there are better options.  Anyone use a different method?

[samhay]

You know you can turn the volume pot down right?
The only downside to adding a series resistor before the volume pot is that you increase the output impedance. This will effect how the circuit interacts with whatever is next in the signal chain. In many cases, it won't be a problem.

[Mark Hammer]

I have a Fender Champion 110 as a sort of bench/try-out-this-pedal amp.  The taper on the volume pots for each channel is terrible, the one on the overdrive channel a bit less so.  This is partly because of how they are wired up.  The schematic shown here is for the CHamp 30, but it's the identical amp/circuit.  You can see that the Clean channel volume simultaneously (and reciprocally) reduces the gain of the op amp AND bleeds off the signal to ground.  Move the wiper to the "right" and C11 is tied directly to ground, while the other side of the wiper has the full 50k (just under 42k when I measured it), going to ground (summing with the existing 7k5 of R20), reducing the gain of that op-amp stage (with the exception of the upper treble that has a direct path to ground via C13). 

To make the zone between 7:00 and 9:00 settings on the volume pot actually usable, without having to completely redesign the circuit and take the whole thing apart, I simply tacked on a resistor between the pot lug going to the wiper, and the lug going to the junction of C11/R23.  This does not change the manner in which the Volume pot alters the gain of the stage, but it affects how much is bled off from C11 at lower volume settings.  I started off with a 47k resistor, but think I may change to 39k, for a little more dialability at lower volumes.

Sometimes, the "solution" to a too-loud problem is using parallel resistors to alter the taper of things.

[blackieNYC]

Word! So many designs with high output volumes. I'd like to put a volume control to somewhere above 9:00 for "unity" gain.  And putting a resistor in series with the output is functionally the same as turning down the pot, in terms of output impedance. Is reducing the pot value the best way to reduce volume but not increase output z?

[anotherjim]

A lot of designs give you room to reduce the pot value and make it back up with a series resistor. Total resistance can be kept the same that way and not change the eq of the output cap.
If a 9v circuit can output that full swing or close, it probably should be padded down. Careful not too much - a max 1v peak to peak ought to be enough, but some designs secretly rely on overdriving the amp - and most players will expect to be able to go noticeably louder than bypassed, even if they don't use it like that.
Example - 100k pot then change to 47k or 50k with 47k or 51k series resistor. Obviously about half with that. Or 22k or 25k pot with 82k - not exactly 100k but won't be noticeable eq change.

I'll just run a 10K-100K resistor to ground right at the output, before my volume pot.

That won't properly work with circuits with low output impedance such as an op-amp - only change the eq of the output cap (less bass).

[Transmogrifox]

Go to Ace Hardware. Buy pack of foam ear plugs.  Install in ears.

30 dB attenuation --- should be enough for most loud pedals.

On a more serious note you might consider an audio transformer for attenuation.  It reduces output impedance while simultaneously attenuating the output level.  Maybe a 10:1 ratio for stuff going 9V peak-peak, but maybe 5:1 more useful as you keep some of the ability to drive a little harder if you need it.

There are a lot of right ways to do this, and in a lot of circumstances a series resistor won't be so bad.  Largest potential consequence of increasing output impedance too much is high frequency roll-off against cable capacitance. 

You can pretty easily get to <10k output impedance by putting a 10k to ground across the pot, and (like your example) 100k in series.  The output impedance is 10k in parallel with 100k in parallel with pot when maxed.  Depending on the pot value it will be dominated by the pot resistance.  If the pot is a smaller value like 10k, then just put in a series resistor.

The output impedance is the output to ground resistance in parallel with the output to source driver resistance.  If you have a 10k from output to ground you will never have more than 10k output impedance.

[blackieNYC]

Looking at some specs (MXR,boss) I see output impedances of 100, 1k, 10k. So there is some variety, but it seems like a series resistor before the pot will be a huge change in output Z, as does a 100k pot in the 10 o'clock position for example, yes?  I mean, it doesn't sound to me like the character of the pedal changes or loses high end content.  I guess this because our next pedal has (optimistically) a 1M input impedance which maintains the well advised 1:10 out/in Z ratio? Which means the series resistor and reduced value pot is a perfectly valid means of attenuation, as long as the total of series resistor plus pot is around 100k or less. Have I got that right?

[Paul Marossy]

Resistor volume divider on the output?

[Transmogrifox]

Looking at some specs (MXR,boss) I see output impedances of 100, 1k, 10k. So there is some variety, but it seems like a series resistor before the pot will be a huge change in output Z, as does a 100k pot in the 10 o'clock position for example, yes?  I mean, it doesn't sound to me like the character of the pedal changes or loses high end content.  I guess this because our next pedal has (optimistically) a 1M input impedance which maintains the well advised 1:10 out/in Z ratio? Which means the series resistor and reduced value pot is a perfectly valid means of attenuation, as long as the total of series resistor plus pot is around 100k or less. Have I got that right?

As I posted before, the equivalent output impedance is the shunt resistance (resistance to ground looking back at the output).

If you have a 10k to ground on the output being fed by a 100k resistor then the output impedance is 10k in parallel with 100k = 9.09k

The series 100k resistor will result in varying degrees of attenuation depending on what you connect to the pedal, but you will never have a high-frequency roll-off lower than f3dB = 1/(2*π*10k*C_cable_capacitance).

If the attenuation is too much then change the 100k to something smaller, but the 100k into 10k should work for most anything because the following stage of most guitar FX is >50k, so you probably won't see a change too much to adjust with the output pot.
100k
--\/\/\/\------------
              |        |
              /        /
       10k \        \  >100k Pot
              /        /``````````````OUTPUT   <---Zo < 10k @ pot max, but dominated by pot in most positions
              \       \
              /       /
              ```|```
               GND

OR

100k
--\/\/\/\------------
              |        |
              /        /
       Rp  \        \  <100k Pot
              /        /``````````````OUTPUT   <---Zo < 10k + amount added by pot
              \       \
              /       /
              ``|``
               GND

Rp*Rpot/(Rp+Rpot) = 10k
   or more simply
Rp = Rpot/(Rpot-10k)

Either way, with 10k shunt to ground as seen looking back into the output you never ADD more than 10k output impedance to whatever you already get from the pot.

[MrStab]

if there are any plain-old inverting op-amps within the circuit, just lower the feedback loop resistance for some attenuation.

similarly, look around for non-inverting amps and lower one of the gain resistors for less amplification, or look around for any voltage dividers that don't noticeably attenuate the signal and abuse them. for an example off the top of my head, in the TS808 (http://gaussmarkov.net/layouts/ts808/ts808-schem.png), you could probably mess with R7 & R8, so long as you compensate the cap for the same roll-off if needed.

be aware of what expectations the circuit has further downstream, though.

[Bishop Vogue]

These are fantastic responses, guys.  Thanks very much.  Judging by the number of views, I was not the only curious one (it's nice to have company).  Although I don't think the poster was serious when he suggested just turning down the volume, I did feel I should explain my concern: if I sell a pedal to someone through a retail outlet and it is WAY louder than the customer expected, at the least they will get a nasty surprise, at worst, they might actually damage something - or so I worry.  At any rate, who wants a pedal they never turn up past 9:00?

Once again, thanks for the great replies.

[samhay]

I was being serious about turning down the volume.
But I'll qualify that. If you are building something designed to push the front end of an amp - i.e. a booster, then the last thing you want to do is not make it loud enough.

We tend to power stuff with 9V, so your signal, assuming no shenanigans with transformers, charge pumps or magic, is not going to be larger than 9V peak-to-peak (it will usually be closer to half this). This is loud, but not blow-up-the-amp loud. Further, if you use a volume pot with a log taper, then you will get at most about 1V peak-to-peak out when the volume pot is at noon. In my opinion, this will quite reasonable.